I have recently been playing around with abelian extensions and I have have found myself playing with Magma and computing conductors of cubic extensions of $F=\mathbb Q(\omega)$, where $\omega$ is a primitive third root of unity. I have stumbled upon Noah Snyder's thesis and more specifically its Proposition 3.6.7, the relevant part of which I will copy here for completeness:
Take any prime $\pi\in F$ lying over $p$ in $\mathbb Q$ relatively prime to $3$. Let $K=F(\sqrt[3]{\pi})$. The finite part of the conductor $\mathfrak f(K/F)$ divides $3\pi$.
The problem I encounter is that this statement seems to be false unless $\pi$ is assumed to be primary. I first confirmed that using Magma, for a specific example of $\pi=3+\omega$:
K<w> := CyclotomicField(3);
p := (3+1*w);
L := ext<K|Polynomial([-p,0,0,1])>;
L :=AbelianExtension(L);
D := Discriminant(L);
C := Conductor(L);
3*p in C;
returns false, as does it for other nonprimary primes I have tried.
It appears that the "fishy" part of the proof in the last part showing $\mathcal O_K\neq\mathcal O_F[\sqrt[3]{\pi}]$. Apart from the typo that $1\pm\sqrt[3]{\pi}\pm\sqrt[3]{\pi^2}$ should be $1\pm\sqrt[3]{\pi}+\sqrt[3]{\pi^2}$, we have a problem that $\sqrt{-3}$ can ramify (and in nonprimary case it always does - $(1\pm\sqrt[3]{\pi})^3$ is divisible exactly once by $1-\omega$). If in the last line we replace $\sqrt{-3}$ by its prime divisor in $\mathcal O_F$, we no longer have an element which necessarily is outside $\mathcal O_K[\sqrt[3]{\pi}]$.
Here come my questions:
Can this argument be salvaged in the primary case? It seems that unfortunately $\sqrt{-3}$ can still ramify in $F$ and even if not, it merely splitting can cause issues as well.
Can we show that if $\pi$ is not primary, then the statement always fails? This amount to showing that in this case $\mathcal O_K=\mathcal O_F[\sqrt[3]{\pi}]$. I tried adapting the argument from $\mathbb Q(\sqrt[3]{n}),n\not\equiv\pm 1\pmod 9$, but couldn't get the argument to work.